The Natural Logarithm as an Integral

Recall the power rule for integrals:

\[\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C , n \neq − 1 .\]

Clearly, this does not work when \(n = −1 ,\) as it would force us to divide by zero. So, what do we do with \(\int \frac{1}{x} d x ?\) Recall from the Fundamental Theorem of Calculus that \(\int_{1}^{x} \frac{1}{t} d t\) is an antiderivative of \(1 / x .\) Therefore, we can make the following definition.

Definition

For \(x > 0 ,\) define the natural logarithm function by

\[\text{ln} x = \int_{1}^{x} \frac{1}{t} d t .\]

For \(x > 1 ,\) this is just the area under the curve \(y = 1 / t\) from \(1\) to \(x .\) For \(x < 1 ,\) we have \(\int_{1}^{x} \frac{1}{t} d t = − \int_{x}^{1} \frac{1}{t} d t ,\) so in this case it is the negative of the area under the curve from \(x \text{to} 1\) (see the following figure).

This figure has two graphs. The first is the curve y=1/t. It is decreasing and in the first quadrant. Under the curve is a shaded area. The area is bounded to the left at x=1. The area is labeled “area=lnx”. The second graph is the same curve y=1/t. It has shaded area under the curve bounded to the right by x=1. It is labeled “area=-lnx”.

Figure 6.75 (a) When \(x > 1 ,\) the natural logarithm is the area under the curve \(y = 1 / t\) from \(1 \text{to} x .\) (b) When \(x < 1 ,\) the natural logarithm is the negative of the area under the curve from \(x\) to \(1 .\)

Notice that \(\text{ln} 1 = 0 .\) Furthermore, the function \(y = 1 / t > 0\) for \(x > 0 .\) Therefore, by the properties of integrals, it is clear that \(\text{ln} x\) is increasing for \(x > 0 .\)

This lesson is part of:

Applications of Integration

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